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d1i1m1o1n [39]
3 years ago
10

If the length of a ramp is increased, what will happen to the input force?

Physics
1 answer:
Art [367]3 years ago
3 0
The input force will increase too because the more the ramp is increasing the more force is being put on the object going down the ramp... hope this helps ^-^
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If light intensity obeyed an inverse square law you would expect to find the intensity of light to decrease as the square of the
Reptile [31]
Inverse square law: \frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2
where 
I_{1} is the intensity at distance 1
I_{2} is the intensity at distance 2
d_{1} is distance 1
d_{2} is distance 2

The inverse squared law state that intensity decreases in inverse proportion to the distance squared. So if light obeyed that rule, it will decreases its intensity as the square of the distance increases.

We can conclude that the correct answer is: true.
4 0
3 years ago
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15 points Plz help me no links
wolverine [178]

Answer:

alternating mountain ranges and valley.

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7 0
3 years ago
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Please help! Will mark Brainliest.
valentinak56 [21]

Answer:

18 Nm

Explanation:

if the correct answer

5 0
3 years ago
The following represents a mass attached to a spring oscillating in simple harmonic motion. X(t) = 4.0 cos(3.0t +0.10) units of
kolbaska11 [484]

Answer:

a) A = 4.0 m , b)   w = 3.0 rad / s , c)  f = 0.477 Hz , d) T = 20.94 s

Explanation:

The equation that describes the oscillatory motion is

          x = A cos (wt + fi)

In the exercise we are told that the expression is

          x = 4.0 cos (3.0 t + 0.10)

let's answer the different questions

a) the amplitude is

         A = 4.0 m

b) the frequency or angular velocity

         w = 3.0 rad / s

c) angular velocity and frequency are related

          w = 2π f

           f = w / 2π

           f = 3 / 2π

           f = 0.477 Hz

d) the period

frequency and period are related

           T = 1 / f

           T = 1 / 0.477

           T = 20.94 s

e) the phase constant

          Ф = 0.10 rad

f) velocity is defined by

          v = dx / dt

         

         v = - A w sin (wt + Ф)

speed is maximum when sine is + -1

         v = A w

          v = 4 3

          v = 12 m / s

g) the angular velocity is

          w² = k / m

          k = m w²

          k = 1.2 3²

          k = 10.8 N / m

h) the total energy of the oscillator is

          Em = ½ k A²

           Em = ½ 10.8 4²

          Em = 43.2 J

i) the potential energy is

           Ke = ½ k x²

for t = 0 x = 4 cos (0 + 0.1)

               x = 3.98 m

j) kinetic energy

           K = ½ m v²

for t = 00.1 ²

    v = A w sin 0.10

    v = 4 3 sin 0.10

    v = 1.98 m / s

3 0
3 years ago
For a body falling freely from rest​ (disregarding air​ resistance), the distance the body falls varies directly as the square o
jasenka [17]

Answer:

The answer to the question is

The object would fall 57.625 m in the first 5 seconds

Explanation:

To solve the question, we note that

the height of fall = 490 ft = ‪149.352‬ m

Time to touch the ground = 7 seconds

We are required to find out how far the object falls in the first 5 seconds

We apply the relation

S = u·t + 0.5×g·t ² = We then have

‪149.352‬ = U×7+0.5*9.81*49 From where u = -13 m/s

Therefore to find how far it falls in the first 5 seconds, we have

-13*5 + 0.5*9.81*25 = 57.625 m

5 0
3 years ago
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